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Factoring Trinomials Where a = 1 In the last section, we discussed factoring trinomials where a ≠ 1. We used the expansion (ac) method where we multiplied the first and last numbers (a and c), and then selected the two factors when added equal the middle number (b). Those two factors were used to rewrite and expand the middle term. In this section we will look at trinomials where a = 1, using a similar procedure. When a = 1, if we multiply the first and last numbers, we will have a product equal to the last number: 1 · c = c. Thus, we only need to select two factors of the last number (c) when added equal the middle number (b). Example 1: Factor: x2 + 2x – 24 Factors of 24: 1,24; 2,12; 3,8; 4,6 Add to get 2. Since the product is -24, one number must be negative: -1+24 = 23 or -24+1 = -23 No -2+12 = 10 or -12+2 = -10 No -3+8 = 5 or -8+3 = -5 No -4+6 = 2 or -6+4 = -2 Yes 2 x – 4x + 6x – 24 Rewrite and expand 2x x2 – 4x + 6x – 24 Split the polynomial in half x(x – 4) + 6(x – 4) Factor the GCF from both sides (x – 4)(x + 6) Notice that we didn’t need to use the expansion method. The final two binomials consist of the factors that add to the middle number. Once we find those two numbers, we can simply write the two binomial factors of the trinomial. Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0) Example 2: Factor: x2 + 9x + 18 Factors of 18: 1,18; 2,9; 3,6 Add to get 9: 1+18 = 19 No 2+9 = 11 No 3+6 = 9 Yes (x + 3)(x + 6) To check the answer, multiply: (x + 3)(x + 6) x(x + 6) + 3(x + 6) x2 + 6x + 3x + 18 x2 + 9x + 18 Correct Example 3: Factor: y2 – 4y + 3 Factors of 3: 1,3 Since the middle term is negative, both numbers must be negative (y – 1)(y – 3) To check the answer, multiply: (y – 1)(y – 3) y(y – 3) – 1(y – 3) y2 – 3y – 1y + 3 y2 – 4y + 3 Correct Example 4: Factor: a2 – 8a – 20 Factors of 20: 1,20; 2,10; 4,5 Add to get -8. Since the product is -20, one number must be negative: -1+20 = 19 or -20+1 = -19 No -2+10 = 8 or -10+2 = -8 Yes (a – 10)(a + 2) Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0) To check the answer, multiply: (a – 10)(a + 2) a(a + 2) – 10(a + 2) a2 + 2a – 10a – 20 a2 – 8a – 20 Correct Example 5: Factor: x2 – 9xy + 14y2 Factors of 14: 1,14; 2,7 Add to get -9. Since the product is positive, both numbers must be negative: -1+(-14) = -15 No -2+(-7) = -9 Yes (x – 2y)(x – 7y) Don’t forget about the variable y To check the answer, multiply: x(x – 7y) – 2y(x – 7y) x2 – 7xy – 2xy + 14y2 x2 – 9xy + 14y2 Example 7: Factor: m2 – mn – 30n2 Factors of 30: 1,30; 2,15; 3,10; 5,6 Add to get -1. Since the product is -30, one number must be negative: -1+30 = 29 or -30+1 = -29 No -2+15 = 13 or -15+2 = -13 No -3+10 = 7 or -10+3 = -7 No -5+6 = 1 or -6+5 = -1 Yes (m – 6n)(m + 5n) Don’t forget the variable n To check the answer, multiply: (m – 6n)(m + 5n) m(m + 5n) – 6n(m + 5n) m2 + 5mn – 6mn – 30n2 m2 – 1mn – 30n2 m2 – mn – 30n2 Correct Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0) Example 6: Factor: x2 + 5x + 8 Factors of 8: 1,8; 2,4 Add to get 5: 1+8 = 9 No 2+4 = 6 No Since no set of factors add to 5, the trinomial cannot be factored. Prime When factoring, it is important not to forget about the GCF. If all terms have a common factor, first factor out the GCF before factoring further. Example 8: Factor: 3z2 – 24z + 45 3(z2 – 8z + 15) Factor out the GCF, 3, first Factors of 15: 1,15; 3,5 Add to get -8. Since the product is positive, both numbers must be negative: -1+(-15) = -16 No -3+(-5) = -8 Yes (z – 3)(z – 5) To check the answer, multiply: 3(z – 3)(z – 5) 3[z(z – 5) – 3(z – 5)] 3[z2 – 5z – 3z + 15] 3[z2 – 8z + 15] 3z2 – 24z + 45 Correct Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)